Find the -intercept(s) and the coordinates of the vertex for the parabola . If there is more than one -intercept, separate them with commas.

Question

y=-x2-4x-4

Answer 1

Some basics first: intercepts are where the curve touches the x-axis and/or the y-axis, so this equation will have both a y-intercept and no more than 2 x-intercepts.  The vertex is where the curve changes direction.  The standard form of the quadratic equation (what you gave us) is y = ax² +bx +c

 

If I'm reading your equation correctly, you gave y = -x² -4x -4.  So, a = -1, b = -4, c = -4

 

y-intercept: when x=0

 y = -0² -4*0 -4

 y = -4

The y-intercept is: (0, -4)

 

vertex: uses the axis of symmetry first...  x= -b/2a  so x = -(-4)/(2*(-1))  which is x = 4/-2 or just  x = -2

So, the x value of the vertex is -2.  If x = -2 (and we just saw on the previous line that it does), then we can substitute that into the equation everywhere there is an x.

y = -(-2)² -4*(-2) -4   Please be careful with the order of operations.

y = -(4)    -4*(-2) -4

y = -4       +8    -4

y = 0

The vertex is at (-2, 0).

 

So, what about the x-intercepts?  There are many different ways to solve for those.  I prefer factoring them, but that's really hard to show in these constraints of this text-only comment box.

 

The Quadratic Formula can find these x-intercepts.  It's actually two formulas in one, so I'll pull them apart.

x= -b ± √(b²-4ac)     The ± means that we will calculate it with the + and then we will calculate it with the -.

                2a             Like I said, there will be up to two intercepts.  This is how we find both.

 

x= -(-4) ± √((-4)²-4(-1)(-4))  Since the b²-4ac part is considered the "Discriminant" and can be useful, I 

            2(-1)                          advise figuring that part separately.

 

x=   4 ± √(16-16)           Again, the Discriminant doesn't always make things so easy, but I'll take it. :)
                -2 

 

x = 4/-2  or x = -2   Let's check this answer by substituting it back into the original equation. y=0

0 = -(-2)² -4*(-2) -4

0 = -4       +8       -4  This does equal zero, so that's a good sign!  The x-intercept is (-2, 0), or if you prefer... the x-intercepts are (-2, 0) and (-2, 0), and if you've been paying attention, this coordinate point is also used as the vertex... ;)  I gave this Quadratic Formula section in case you were asking this to help you with other questions.  For inquiring minds, because the "a" value is negative, this parabola (what you get when you graph a quadratic equation) will open down and look like a hill.

Answer 2

The x-intercept of an equation is the x value (or values) that makes the y value equal to 0. Graphically, it is the point (or points) where the graph crosses the horizontal x-axis.

 

The vertex of a parabola is the rounded "tip" of the curve, which is either the lowest point of the curve if the parabola opens up or the highest point of the curve if the parabola opens down.

 

You have different options for solving this problem.

 

1) Quadratic Formula
Hopefully, you know that equations of this form, 0 = ax² + bx + c, can solved with the quadratic formula:

x = (-b + SquareRoot(b2 - 4ac))/2a or x = (-b - SquareRoot(b2 - 4ac))/2a

 

Of course, SquareRoot(b² - 4ac) can have 1, 2, or no real value.

 

For example SquareRoot(0) = 0 has 1 value. SquareRoot(25) = 5 or -5 has 2 values. But SquareRoot(-25) has no real values because there is no real number that when multiplied by itself gives a negative answer.

 

So you can use the quadratic formula to solve for x and that will give you either 1, 2, or no x-intercepts.

 

The vertex will always be halfway between the two x-intercepts if there are two x-intercepts, or it will be the x-intercept if there is only one x-intercept. If there are no real x-intercepts, than you must find the x-intercept some other way.

 

2) Completing The Square

You may have learned the method of completing the square, which is the concept that an equation of the form y = ax² + bx + c can be re-written in the form y = (x - h)² + k.

 

As an example, I'll complete the square for y = x² - 2x + 4 with the following steps

 

y + (2/2)² = x² - 2x + (2/2)² + 4  // add the square of half of the coefficient on the x term to both sides

y +    1²   = x2 - 2x +     1²  + 4

y +    1     = x2 - 2x +     1    + 4

y +    1     = (x - 1)²             + 4  // factor the terms with x and the term you added

y  + 1  - 1 = (x - 1)²      +  4  - 1 // now subtract the term you added to both sides

y              = (x - 1)²      + 3

 

So now you have an equavilent representation of y = x² - 2x + 4 = (x - 1)² + 3

 

But you'll note that this form is just like y = x² but shifted 1 unit to the right and up 3. So since y = x² has its vertex at (0,0), y = x² - 2x + 4 has its vertex shifted 1 unit to the right and up 3 units to (1,3).

 

You can follow the same steps to complete the square for y = -x² - 4x - 4 to find the vertex for that parabola.

 

3) Graphically

You cannot get an exact answer with a graphical procedure, but making a table of values to sketch the curve on graph paper can help you visualize the problem. For example:

 

(x, y)

(-3, -(-3)² - 4(-3) - 4) = (-3,  -1)

(-2, -(-2)² - 4(-2) - 4) = (-2,   0)

(-1, -(-1)² - 4(-1) - 4) = (-1,  -1)

( 0,    -0²  -  4(0) - 4) = ( 0,   -4)

( 1,    -1²  -  4(1) - 4) = ( 1,   -9)

( 2,    -2²  -  4(2) - 4) = ( 2, -16)

( 3,    -3²  -  4(3) - 4) = ( 3, -25)

 

Start connecting the points and try to identify the x-value or values at which the graph crosses the x-axis. Also try to find the tip of the curve, which is called the vertex.

 

Even without plotting, you might recognize that y = -x² - 4x - 4 is the equation of a parabola opening downwards, so you know a parabola can have 1, 2, or no x-intercepts.

 

NOTE: The difficult thing about plotting is that you don't know which x values to use in your table of values and you cannot rely on your hand-drawn plots to find the exact answer, which might fall in-between whole numbers.